Multivariable: Derivatives of Polar Functions, Integrals of Polar Functions.

August 4, 2010

This is going to be a short post about basic differentiating and integrating of multivariable functions.  Remember, just because something is polar doesn’t mean that it’s not fun! But, if something is fun, then it rarely is polar.  Just saying.

 

Differentiating.

Obviously, if you want to take \frac{dr}{d\theta}, then you differentiate normally.  But say we have some function r(\theta) and we want to find \frac{dy}{dx}; note that this is slightly different from polar differentiating, and, in fact, will give us the “slope” of the polar plot if we look at it as if it were in rectangular coordinates.

Let’s derive this.  Okay, so, what do we know about y and x in terms of polar coordinates?  Well,

y = r\sin(\theta)

x = r\cos(\theta)

so, that’s nice.  Now, can we manipulate \frac{dy}{dx} to give us something nice?  Let’s apply the chain rule:

\displaystyle \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}

Now, notice that \frac{dy}{d\theta} is saying, “Take the derivative of the function y with respect to \theta.”  What is this?  Well, y = r\sin(\theta), and since r may be a function of \theta (indeed, it usually is written this way), we have

\displaystyle \frac{dy}{d\theta} = \frac{dr}{d\theta}\sin(\theta) + r\cos(\theta)

by applying the product rule, and similarly,

\displaystyle \frac{dx}{d\theta} = \frac{dr}{d\theta}\cos(\theta) - r\sin(\theta)

which allows us to simplify the expression above to:

\displaystyle \frac{dy}{dx} = \frac{\frac{dr}{d\theta}\sin(\theta) + r\cos(\theta)}{\frac{dr}{d\theta}\cos(\theta) - r\sin(\theta)}

This may not look like the simpliest expression, but it is much easier than converting a function from polar form back into rectangular form and differentiating.

Let’s take an example.  Let’s take the function r(\theta) = 3\cos(2\theta), below.

image

Now, look at this.  It looks pretty cool, right?  Yeah, let’s find \frac{dy}{dx} of this, and see what’s what.  First, let’s just calculate \frac{dr}{d\theta} so we can replace it into the equation for the rectangular derivative.

\frac{dr}{d\theta} = -6\sin(2\theta)

\displaystyle \frac{dy}{dx} = \frac{-6\sin(2\theta)\sin(\theta) + 3\cos(2\theta)cos(\theta)}{-6\sin(2\theta)\cos(\theta) - 3\cos(2\theta)\sin(\theta)}

and, now, I’ll be the first to admit that this is really an ugly formula.  But, for now, let’s not reduce it, and let’s just plug in some things to see if we get something nice.  First notice that at \theta = \frac{\pi}{2} (this is the very bottom of the graph) we note that we should have a horizontal tangent line.  In fact, plugging in our value for \theta, we find that our derivative gives us 0!  Nicely done, folks.

Let’s just do one more case, just to show we aren’t makin’ stuff up and everything should work out as expected.  Suppose we have a circle with radius 2; that is, in polar form, we have a constant radius: r(\theta) = 2.

image

In rectangular form, what is this?  It’s just x^{2} + y^{2} = 4, right?  Well, let’s see if we get the same derivatives doing this both ways.  First,

\frac{dr}{d\theta} = 0

So that our above formula gives us

\displaystyle \frac{dy}{dx} = \frac{\cos(\theta)}{-\sin(\theta)} = -\cot(\theta)

which is certainly nice.

Now, let’s use implicit differentiation to solve this rectangular-wise.

\frac{dx}{dx}2x + \frac{dy}{dx}2y = 0

which implies that

\displaystyle \frac{dy}{dx} = \frac{-x}{y}

which, if we recall how to put this into polar form, is

\displaystyle \frac{dy}{dx} = \frac{-r\cos(\theta)}{r\sin(\theta)} = -\cot(\theta)

which is exactly what we wanted!  Good.

 

Integrating.

To begin, let’s describe what we’ll be doing to integrate in polar coordinates.  First, let’s think of what we might do to find an area of a circle.  Let’s split it up into pizza pie slices, and then take the area of those: these are sectors, right?  So, how big of a chunk is the sector of the circle?  Let’s suppose that \theta is in radians.

\frac{\theta}{2\pi}

and so, the total area of the sector is

\displaystyle \frac{\theta}{2\pi}\pi r^{2} = \frac{\theta}{2}r^{2}

as we know from geometry, where \theta is the degree of each slice.  What is the degree of each slice?  Well, at this point, since it’s a circle, it doesn’t really matter, but let’s try to make them really small.  In fact, let’s make it so that they are infinitely small — in other words, let’s make them into d\theta‘s.  How do we sum up infinitely small things?  Integrate!

\displaystyle \int_{0}^{2\pi}\frac{1}{2}r^{2}d\theta

will give us the area of the circle for some constant radius.  Now, how about if we were to change the radius?  Well, since we have d\theta, it’s almost like we’re integrating little bits of concentric circles.

polarint

On this graph, for example, this part of the sector (VERY exaggerated), we’re pretending that this part of the graph is a circle.  As we move, the circle gets a little bigger or smaller.  So, basically, if we have some equation r(\theta) we can integrate it by replacing r for r(\theta).  We then get

\displaystyle \int_{a}^{b} \frac{1}{2}(r(\theta))^{2}d\theta

for the area under a polar curve from \theta = a to \theta = b.

Now, be cautious here: if the graph “overlaps” itself, you’ll compute a much larger area than you wanted!  Let’s give an example of this.

The function r(\theta) = 2, as you recall, is a circle with radius 2, and we have

\displaystyle \int_{0}^{2\pi} d\theta = 2\pi

but, if we were to take

\displaystyle \int_{0}^{4\pi} d\theta = 4\pi

so what gives?  It’s because in the latter one, we went around the circle twice. Yeah, so it’s kind of like looking at a spiral, and the first one goes around once and the second one goes around twice.

Now, let’s just look at slightly more difficult example.  What about r(\theta) = 3cos(\theta)?

image

Well, let’s integrate this.

\displaystyle \int_{a}^{b} \frac{9}{2}\cos^{2}(\theta)d\theta = \frac{9}{2}\int_{a}^{b}\cos^{2}(\theta)d\theta

and, as you know, the power-reducing formula for cosine squared gives us:

= \displaystyle \frac{9}{2}\int_{a}^{b}\frac{1 + \cos(\theta)}{2}d\theta

= \displaystyle \frac{9}{2}|\frac{\theta + \sin(\theta)}{2}|_{a}^{b}

= \displaystyle \frac{9}{4}|\theta + \sin(\theta)|_{a}^{b}

but, now, what are our limits of integration?  Well, if we look at the graph, it’s going from 0 to \pi.

So, this reduces to

\displaystyle \frac{9}{4}|\theta + \sin(\theta)|_{0}^{\pi}

= \displaystyle \frac{9}{4}((\pi + 0) - (0 + 0)) = \frac{9\pi}{4}

But how can we check this?  Notice that this is, in fact, a circle, and that the farthest point is when cosine is equal to 1; in other words, this is a circle of diameter 3.  This means the radius is equal to \frac{3}{2} and, therefore, the area is \frac{9\pi}{4}.  I’ll hold for gasps and applause.

 

Observations and Notes.

This is essentially it.  The hardest part is really finding out what the upper and lower limits of this type of equation is.  For this, my best solution has been: use wolfram alpha.  Seriously.  I’m pretty lazy.

Since this type of thinking will generalize to, say, spherical and cylinderal coordinates, it would be easy to derive similar methods of differentiation and integration for these representations.

Maybe we’ll do these.  Maybe.

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